a: \(A=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

\(ĐK:a>0;a\ne1;a\ne4\\ a,A=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,A>0\Leftrightarrow\sqrt{a}-2>0\Leftrightarrow a>4\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

Câu 2: 

Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)

\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)

\(=1-a\)

Câu 1: 

Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)

\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)

\(=1\)

Sửa đề: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

a) Ta có: \(P=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a-1}+\dfrac{\sqrt{a}}{\sqrt{a}+1}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

\(=\left(\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

\(=\dfrac{\sqrt{a}+1-2\sqrt{a}+a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{\sqrt{a}}{\sqrt{a}+1}\)

\(=\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

\(=\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

b) Để \(P< \dfrac{1}{2}\) thì \(P-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-1}{\sqrt{a}}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{a}-1\right)}{2\sqrt{a}}-\dfrac{\sqrt{a}}{2\sqrt{a}}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{a}-2-\sqrt{a}}{2\sqrt{a}}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{a}-2}{2\sqrt{a}}< 0\)

mà \(2\sqrt{a}>0\forall a\) thỏa mãn ĐKXĐ

nên \(\sqrt{a}-2< 0\)

\(\Leftrightarrow\sqrt{a}< 2\)

hay a<4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)

Vậy: Để \(P< \dfrac{1}{2}\) thì \(\left\{{}\begin{matrix}0< a< 4\\a\ne1\end{matrix}\right.\)

ĐKXĐ: \(-1\le a< 1\); \(a\ne0\)

\(P=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}^2}{\sqrt{\left(1-a\right)\left(1+a\right)}-\sqrt{1-a}^2}\right).\left(\sqrt{\dfrac{\left(1-a\right)\left(1+a\right)}{a^2}}-\dfrac{1}{a}\right)\)

\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\sqrt{\dfrac{\left(1-a\right)\left(1+a\right)}{a^2}}-\dfrac{1}{a}\right)\)

\(=\left(\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)

\(=\dfrac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{1+a-\left(1-a\right)}.\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)

\(=\left(\dfrac{1+\sqrt{\left(1-a\right)\left(1+a\right)}}{a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{\left|a\right|}-\dfrac{1}{a}\right)\)

- Với \(a>0\)

\(\Rightarrow P=\dfrac{\left(\sqrt{1-a^2}+1\right)\left(\sqrt{1-a^2}-1\right)}{a^2}=\dfrac{1-a^2-1}{a^2}=-1\)

- Với \(a< 0\)

\(\Rightarrow P=-\dfrac{\left(1+\sqrt{1-a^2}\right)^2}{a^2}\)

\(B=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)

\(=2+\dfrac{1}{\sqrt{a}}\cdot\dfrac{2a+2}{\sqrt{a}+1}\)

\(=\dfrac{2a+2\sqrt{a}+2a+2}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{4a+2\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

Ta có: \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=2+\dfrac{2a+2}{\sqrt{a}}\)

\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)

\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}\)

\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)